Newtons Universal Law of Gravitation This cartoon mixes

Newton’s Universal Law of Gravitation This cartoon mixes 2 legends: 1. The legend of Newton, the apple & gravity which led to The Universal Law of Gravitation. 2. The legend of William Tell & the apple.

• It was very SIGNIFICANT & PROFOUND in the 1600's when Sir Isaac Newton first wrote Newton's Universal Law of Gravitation! • This was done at the age of about 30. It was this, more than any of his other achievements, which caused him to be well-known in the world scientific community of the late 1600's.

Newton's Universal Law of Gravitation • Newton used this law, along with Newton's 2 nd Law (his 2 nd Law!) plus Calculus, which he also (co-) invented, to PROVE that The orbits of the planets around the sun MUST be ellipses. • For simplicity, we assume in the following that these orbits are circular.

• The topic of Gravitation fits THE COURSE THEME OF NEWTON'S LAWS OF MOTION • Newton used his Gravitation Law & his 2 nd Law in his analysis of planetary motion. His prediction that planetary orbits are elliptical is in excellent agreement with Kepler's analysis of observational data & with Kepler's empirical laws of planetary motion.

• When Newton first wrote the Universal Law of Gravitation, it was the first time, anyone had EVER written a theoretical expression (physics in math form) & used it to PREDICT something in agreement with observations! • For this reason, Newton's formulation of his Universal Gravitation Law is considered BEGINNING OF THEORETICAL PHYSICS. to be the

The Universal Law of Gravitation • Also gave Newton his major “claim to fame”. After this, he was considered to be a “major leader” in science & math among his peers. In modern times, this, plus the many other things he did, have led to the consensus that Sir Isaac Newton was the GREATEST SCIENTIST WHO EVER LIVED

Newton’s Universal Law of Gravitation • This is an EXPERIMENTAL LAW describing the gravitational force of attraction between 2 objects. • Newton’s reasoning: The Gravitational Force of Attraction between 2 large objects (Earth - Moon, etc. ) is the SAME force as the attraction of objects to the Earth.

Newton’s Universal Law of Gravitation The Gravitational Force of Attraction between 2 large objects (Earth - Moon, etc. ) is the SAME force as the attraction of objects to the Earth. Apple Story: • This is likely not a true historical account, but the reasoning discussed there is correct. This story is probably legend rather than fact.

Newton’s Question: • If the force of gravity is being exerted on objects on Earth, what is the origin of that force? • Newton realized that this force must come from the Earth. • He further realized that this force is what keeps the Moon in its orbit.

• The gravitational force on you is half of a Newton’s 3 rd Law Pair • Earth exerts a downward force on you, & you exert an upward force on Earth. When there is such a large difference in the 2 masses, the reaction force (force you exert on the Earth) is undetectable, but for 2 objects with masses closer in size to each other, it can be significant. This must be true from Newton’s 3 rd Law!

Always a Newton’s 3 rd Law Pair! Newton’s Reasoning • The Force of Attraction between 2 small masses is the same as the force between Earth & Moon, Earth & Sun, etc.

• By observing planetary orbits & analyzing other’s observations, Newton concluded that the gravitational force decreases as the inverse square of the distance r between the 2 masses. Newton’s Universal Law of Gravitation: • “Every particle in the Universe attracts every other particle in the Universe with a force that is proportional to the product of the masses & inversely proportional to the square of the distance between them”

Newton’s Universal Law of Gravitation: • “Every particle in the Universe attracts every other particle in the Universe with a force that is proportional to the product of their masses & inversely proportional to the square of the distance between them: F 12 = -F 21 [(m 1 m 2)/r 2] F 12 must = – F 21 from Newton’s 3 rd Law! • The direction of this force is along the line joining the 2 masses.

• The FORCE between 2 small masses has the same origin (Gravity) as the FORCE between the Earth & the Moon, the Earth & the Sun, etc. From Newton’s 3 rd Law!

Newton’s Universal Gravitation Law • This force is written as: G a constant G Universal Gravitational Constant • G is measured & is the same for ALL objects. G must be small!

G Universal Gravitational Constant • G is measured & is the same for ALL objects. G must be small! • Measurement of G in the lab is tedious & sensitive because it is so small. – First done by Cavendish in 1789. Cavendish Measurement Apparatus

G Universal Gravitational Constant • Modern version of the Cavendish experiment: Two small masses are fixed at ends of a light horizontal rod. • Two larger masses are placed near the Cavendish smaller ones. The angle of rotation is Measurement nd measured. Use Newton’s 2 Law to Apparatus get the vector force between the masses. Relate to the angle of rotation & then extract G.

G = Universal Gravitational Constant • Measurements find, in SI Units: • The force given above is strictly valid only for: – Very small (point) masses m 1 & m 2 – Uniform spheres • For other objects: Need integral calculus!

• The Universal Law of Gravitation is an example of an inverse square law – The force magnitude varies as the inverse square of the separation of the particles • The law can be expressed in vector form. • The negative sign means it’s an attractive force! Aren’t we glad it’s not repulsive?

Comments F 12 Force exerted by particle 1 on particle 2 F 21 Force exerted by particle 2 on particle 1 Its also true that F 21 = - F 12 • This tells us that the forces form a Newton’s 3 rd Law action-reaction pair, as expected. • The negative sign in the vector equation tells us that particle 2 is attracted toward particle 1

More Comments • Gravity is a “field force” that always exists between 2 masses, regardless of the medium between them. The gravitational force decreases rapidly as the distance between the 2 masses increases. (This is an obvious consequence of the inverse square law)

Solar System Data (needed for calculations) See also Table 5. 1, page 147 or our book

Solar System Data

Example Gravitational Force Between 2 People • A m 1 = 50 -kg person & m 2 = 70 -kg person are sitting on a bench close to each other. • Estimate the magnitude of the gravitational force each exerts on the other.

Example Gravitational Force Between 2 People • A m 1 = 50 -kg person & m 2 = 70 -kg person are sitting on a bench close to each other. • Estimate the magnitude of the gravitational force each exerts on the other. • Solution: Suppose r = 1. 0 m. • The force is then: F = (Gm 1 m 2)/(r 2) = (6. 67 10 -11)(50)(70)/(1)2 F 2. 33 10 -7 N • Too small to be noticed!

Example: Spacecraft at 2 r. E • A spacecraft of mass m = 2000 kg at an altitude of twice the Earth radius. Numbers from tables: Earth, Mass ME m Earth Radius: r. E = 6320 km Mass: ME = 5. 98 1024 kg FG = G(m. ME/r 2) At r = 2 r. E FG = G[m. ME/(2 r. E 2 )] = 4900 N

Example: Force on the Moon • Find the net force on the Moon due to the gravitational attraction of both the Earth & the Sun, assuming they are at right angles to each other. ME= 5. 99 1024 kg, MM= 7. 35 1022 kg MS = 1. 99 1030 kg r. ME = 3. 85 108 m, r. MS = 1. 5 1011 m F = FME + FMS (A vector sum!)

F = FME + FMS (A vector sum!) FME = G [(MMME)/(r. ME)2] = 1. 99 1020 N FMS = G [(MMMS)/(r. MS)2] = 4. 34 1020 N F = [ (FME)2 + (FMS)2](½) = 4. 77 1020 N tan(θ) = 1. 99/4. 34 θ = 24. 6º

Gravitational Force Due to a Mass Distribution • It can be shown with integral calculus that: The gravitational force exerted by a spherically symmetric mass distribution of uniform density on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at the center. • So, assuming that the Earth is such a sphere, the gravitational force exerted by the Earth on a particle of mass m on or near the Earth’s surface is FG = G[(m. ME)/r 2] ME Earth Mass, r. E Earth Radius

FG = G[(m. ME)/r 2] ME Earth Mass, r. E Earth Radius • Similarly, to treat the gravitational force due to large spherical shaped objects, we can show with calculus, that: 1. If a (point) particle is outside a thin spherical shell, the gravitational force on the particle is the same as if all the mass of the sphere were at center of the shell. 2. If a (point) particle is inside a thin spherical shell, the gravitational force on the particle is zero. So, we can model a sphere as a series of thin shells. For a mass outside any large spherically symmetric mass, the gravitational force acts as though all the mass of the sphere is at the sphere’s center.

Vector Form of the Universal Gravitation Law In vector form: • The figure at the right gives the directions of the displacement & force vectors. • If there are many particles, the total force is the vector sum of the individual forces:

Example: Billiards (Pool) • 3 billiard (pool) balls, masses m 1 = m 2 = m 3 = 0. 3 kg on a table as in the figure. Triangle sides: a = 0. 4 m, b = 0. 3 m, c = 0. 5 m. • Calculate the magnitude & direction of the total gravitational force F on m 1 due to m 2 & m 3. • Note: Gravitational force is a vector, so we have to add the vectors F 21 & F 31 to get the vector F (using the vector addition methods of earlier). F = F 21 + F 31

Example: Billiards (Pool) • 3 billiard (pool) balls, masses m 1 = m 2 = m 3 = 0. 3 kg on a table as in the figure. Triangle sides: a = 0. 4 m, b = 0. 3 m, c = 0. 5 m. • Calculate the magnitude & direction of the total gravitational force F on m 1 due to m 2 & m 3. • Note: Gravitational force is a vector, so we have to add the vectors F 21 & F 31 to get the vector F (using the vector addition methods of earlier). F = F 21 + F 31 • Using components: Fx = F 21 x + F 31 x = 0 + 6. 67 10 -11 N Fy = F 21 y + F 31 y = 3. 75 10 -11 N • So, F = [(Fx)2 + (Fy)2]½ = 3. 75 10 -11 N tanθ = 0. 562, θ = 29. 3º

Gravity Near the Earth’s Surface The Gravitational Acceleration g and The Gravitational Constant G

G vs. g • Obviously, it’s very important to distinguish between G and g! – They are clearly very different physical quantities! G Universal Gravitational Constant • It’s the same everywhere in the Universe G = 6. 673 -11 10 2 2 N∙m /kg ALWAYS at every location anywhere g The Acceleration Due to Gravity • g = 9. 80 m/s 2 (approximately!) on the Earth’s surface. g varies with location

g in terms of G • Consider a mass m on Earth’s surface: m. E = mass of Earth (say, known) r. E = radius of Earth (known) m = mass of the object (known) • Assume that the Earth is a uniform, perfect sphere. • The weight of m is FG = mg • The Gravitational force on m is FG = G[(mm. E)/(r. E)2] • Setting these equal gives: • All quantities on the right are measured! m m. E g = 9. 8 m/s 2

• Using the same process, we can m “Weigh” the Earth! (Determine it’s mass). • On Earth’s surface, equate the usual weight of mass m to the m. E Newton’s Gravitation Law form for the gravitational force: All quantities on the right are measured! • Knowing g = 9. 8 m/s 2 & the radius of the Earth r. E, the mass of the Earth m. E can be calculated: m. E 6 1024 kg

Effective Acceleration Due to Gravity • Consider the acceleration due to gravity at a distance r from Earth’s center. • Write the gravitational force as: FG = G[(m. ME)/r 2] mg ( “Effective Weight” of m) g Effective acceleration due to gravity. SO : g = G(ME 2 )/r ME

Altitude Dependence of g • If an object is some distance h above the Earth’s surface, r becomes RE + h. Again, set the gravitational force equal to mg : G[(m. ME • This gives: )/r 2] mg ME • This shows that g decreases with increasing altitude • As r ® , the weight of the object approaches zero Example, g on Mt. Everest

Altitude Dependence of g Lubbock, TX: Altitude: h 3300 ft 1100 m g 9. 798 m/s 2 Mt. Everest: Altitude: h 8. 8 km g 9. 77 m/s 2

Example: Effect of Earth’s Rotation on g • Assume that Earth is a perfect sphere. • Determine how its rotation affects the value of g at the equator compared to its value at the poles. • Let towards the center of the Earth be positive y. nd 2 Newton’s Law is: ∑F = ma. R = mg – W so W = mg – ma. R = mg – m(v 2/r)

Example: Effect of Earth’s Rotation on g Newton’s 2 nd Law: ∑F = ma. R = mg – W so W = mg – ma. R W = mg – m(v 2/r) • At the pole, v = 0, so W – mg = 0 & W = mg • At the equator, the centripetal acceleration is a. R = [(v 2)/(r. E)] • So, the effective weight of a mass m there is: W = mg – ma. R = mg - m[(v 2)/(r. E)] = mg

Example: Effect of Earth’s Rotation on g • Effective weight of mass m at equator: W = mg – ma. R = mg - m[(v 2)/(r. E)] = mg Period = T = 1 day = 8. 64 104 s • So the speed v there is v = (2πr. E)/T = 4. 64 102 m/s • The centripetal acceleration at the equator is a. C = [(v 2)/(r. E)] = 0. 037 m/s 2 • Effective gravitational acceleration at equator: g = g - [(v 2)/(r. E)] = 9. 8 - 0. 037 = 9. 763 m/s 2

“Weighing” the Sun! • We’ve “weighed” the Earth, now lets “Weigh” the Sun!! (Determine it’s mass). • Assume: Earth & Sun are perfect uniform spheres & Earth orbit is a perfect circle (actually its an ellipse). • Note: For Earth: Mass ME = 5. 99 1024 kg The orbit period is T = 1 yr 3 107 s The orbit radius r = 1. 5 1011 m The orbit velocity v = (2πr/T), = v 3 104 m/s

“Weighing” the Sun! • Note: For Earth: Mass ME = 5. 99 1024 kg The orbit period is T = 1 yr 3 107 s The orbit radius r = 1. 5 1011 m The orbit velocity v = (2πr/T), = v 3 104 m/s • The Gravitational Force between the Earth & the Sun: FG = G[(MSME)/r 2] • Circular orbit Centripetal acceleration • Newton’s 2 nd Law for the Earth gives: ∑F = FG = MEac = ME(v 2)/r OR: G[(MSME)/r 2] = ME(v 2)/r. • If the Sun mass is unknown, solve for it: MS = (v 2 r)/G 2 1030 kg 3. 3 105 ME

Gravitation and the Moon’s Orbit • The Moon follows an approximately circular orbit around the Earth • There is a force required for this motion • Gravity supplies the Centripetal Force

The Moon’s Motion • We’ve assumed that the Moon orbits a “fixed” Earth – This is a good approximation – It ignores Earth’s motion around the Sun • The Earth and Moon actually both orbit their center of mass – We can think of the Earth as orbiting the Moon – The circle of the Earth’s motion is very small compared to the Moon’s orbit